Do X Then Do X Again

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The Section of Mathematics Pedagogy

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The Product of Two Linear Functions
each of which is Tangent to the Product Function

by

James Due west. Wilson
Academy of Georgia

and

David Barnes
University of Missouri

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This may take been an attempt to write a newspaper with a longer title than the paper itself. In fact, this "paper" is a discussion of our examining a detail trouble that having tools like function graphers bachelor might make possible different approaches.

The problem is:

Find two linear functions f(x) and chiliad(ten) such that the product
h(x) = f(x).g(ten) is tangent to each.

This problem was posed by a group of teachers during a workshop in which the apply of function graphers was existence explored. Our analysis is presented as a sort of stream of consciousness account of how 1 might explore the problem with the tools at hand. In fact, we came up with 2 different streams of consciousness and then nosotros accept two senarios that are parallel in that they cover alternative approaches to the problem. The senarios stand for a blended of several discussions of the problem with teachers, students, and colleagues. Note, the goal here is using this problem context, not only to solve the problem posed, but to understand the concepts and procedures underlying the trouble.

Function graphers are available for almost any computing platform or graphic calculator. Such tools make it possible to look at new topics in the mathematics curriculum or to look at current topics in different means. This problem has some elements of each. In general information technology would not be included in the schoolhouse curriculum, only there is no reason information technology should not. Further, the use of technological tools to examine visualizations of the functions makes for a unlike approach to the trouble.

Is it possible to discover 2 linear functions, f(x) = mx + b and g(10) = nx + c, such that the function h(x) = f(x).thousand(x) is tangent to each. A traditional approach would begin with algebraic manipulation. This is useful because information technology keeps the students occupied, simply what do they larn from it? Clearly, h(x) = (mx + b)(nx + c) is a polynomial of caste two and h(ten) has two roots. The respective roots are when f(ten) = 0 and g(x) = 0. This means the graph of h(10) crosses the x-axis at the aforementioned two points equally f(10) and m(10). Thus, if at that place are points of tangency and so they must occur at these common points on the x-centrality. Experienced students, very bright students, and good problem solvers could whittle this information and a lot more information out of this algebraic assay. Novice students would be more than tentative.

Senario One

Lets open up the role grapher and explore with some specific f(ten), one thousand(x), and the resulting h(10). Lets try

The graphs on the aforementioned ready of axes are

For some novices, seeing the graph of the product h(10) = (3x + ii)(2x+ane) and the graphs of the two directly lines from the factors on the same coordinate axes provides a new feel. This particular graph has one of the two lines "close" to being tangent to the product curve but the other ane is not close. How could the motion-picture show be changed?

One idea is to spread the two lines so that i has negative gradient. Endeavour

The graphs are

This is ameliorate? What tin be observed? How can the graph of h(10) be "moved down?" What if the graphs of f(x) and g(x) had smaller y-intercepts? Effort

The graphs are

Nonetheless not as well practiced but at to the lowest degree the graph of h(10) was "moved downward."

Try smaller y-intercepts, such as

and issue in graphs of

These graphs seem close, but clearly the line with negative slope is not tangent to the graph of h(10). Looking back over the sequence of graphs (and perhaps generating some others) the graph of h(x) always has a line of symmetry parallel to the y-axis. It seems that the pair of tangent lines volition have to take this same symmetry. How? Try making the slopes iii and -three. The functions are

and the graphs are

A zoom to the correct hand side of the graph with requite

showing tangency has non been achieved. A zoom to the left hand side shows a like problem.

One could attempt adjusting the y-intercepts. In fact, if the y-intercepts were equal, the y-centrality would be the line of symmetry. Try

The resulting graphs are

Worse. Effort

The graphs are

A zoom to the left shows

and to the right shows

The upshot seems on target. It remains to confirm f(10) and thou(10) each share exactly i betoken in common with h(10). Once again the tradition is to exercise so algebraically, simply information technology might exist instructive to await at some graphs of h(ten) - f(x) and h(x) - grand(x), such as the following:

Can we immediately generate graphs of other f(x), g(x), and h(10) satisfying the conditions of the problem? Reviewing the graphs and the strategy, its seems that the slopes of the lines can vary, the but condition existence that they are m and -g. So, a simpler instance might exist to permit the slopes be 1 and -one, giving

and

It is likewise of interest to see both of the solutions on the aforementioned graph:

Other solutions could be generated past making some other vertical line the axis of symmetry. Indeed substituting

gives the graphs

for which the equations simplify to

First consider the graphs of f(x) and chiliad(x) and try to sketch in h(x). The graph is

What do these lines tell you lot about the parabola? What points do yous know the bend volition go through? Why? What causes it to open the fashion information technology does? Now let's add the graph of the parabola and compare information technology with our sketch.

How is information technology like our sketch? How is it dissimilar? Is there anything nosotros should observe or consider?

Information technology appears that all iii graphs seem to intersect at i on the y axis. Lets zoom for a closer look.

Maybe irresolute one of the functions will assistance with the caption.
Consider

The three functions no longer intersect at 1 on the y-axis. However, the changed function, f(x), does intersect the curve at its y-intercept.

When g(x) = ane the parabola intersects f(x). Is the contrary true? Lets graph and run across.

It seems that if f(x) = 1 then h(x) = g(x) and if k(x) = 1 so h(x) = f(x). Lets test this by trying to generate h(x) from a new f(ten) and yard(x). Permit

Then add together the sketch h(x) and compare with . . .

.

In this process we seem to have also noticed that the lines and the parabola intersect at the points when the lines cross the x-axis. Why would this exist truthful?

At present the goal is to become ane line tangent to the parabola. The function g(x) is close to being tangent. If we could merely get the two points to slid together then they would go 1 point -- the betoken of tangency. (If a line intersects a parabola in exactly one point, what is true about the line?)

Since f(x) takes on a value of i when x = one, so lets try to change g(10) so that yard(ane) = 0. Lets see we could change the slope or change thou(x)'due south position up and down.

Lets try them both.

To alter gradient, 1000(x) = -2x + ii and test to see if g(1) = -2(i) + 2 = 0

Or modify position, 1000(x) = -3x + 3 and test m(1) = -3(1) + three = 0

This seems to imply that f(x) and h(x) are tangent at the f(x) and h(ten)'s common root, if the function g(ten) takes on the value 1 at this root. Or in other words if f(a) = 0 and 1000(a) = 1 then f(10) is tangent to h(x) at a.

What would we have to practise to become both f and g tangent to h? That would hateful that when f(x) = 0, then g(ten) = ane; and when g(10) = 0, then f(10) = i. Lets showtime start with an easy office for f and and then try to generate a thou(x) which satisfies what we desire. Lets begin with f(x) = 10.

Now when f(10) = 0, then chiliad(ten) should take a value of 1. In other words if f(0) = 0, then one thousand(0) = 1. Likewise, when g(ten) = 0, f(x) should take a value of one. Since f(1) = ane then thou(1) = 0. We need a our linear part chiliad(x) to go through (0,1) and (one,0). And so our k(10) = -ten +i. Lets graph it to check.

That looks right! Now, add the graph of the production so test information technology to see if the curves are tangent.

That looks good. (What is the coordinates of the vertex?) Lets zoom in at the roots.

and .

This seems to be a useful direction. Does it work on the previous trouble? When we left off, f(x) = x and g(x) = -3x + 3. Can nosotros utilize our technique to find a dissimilar f(ten) that works for g(x) to produce h(x) = f(ten).g(ten) with f(x) and yard(x) each tangent to h(x)? We accept the following graph.

Since g(1) = 0, then f(i) = 1 and since 1000(two/3) = 1 then f(2/iii) = 0 will be necessary. So f(x) contains the points (1,1) and (2/3, 0). Try g(x) = -3x - 2.

Zoom in for a closer look.

What are the coordinates of the upper vertex of the triangle? What are the coordinates of the vertex of the parabola? Are the lines actually tangent to the parabola? How might this be proved? Where else could the function f(x) possibly accept on the same value equally h(ten) if h(x) = f(x).thou(x)? And how tin we interpret this on the graphs?

Summary

Many issues are hidden in these composite accounts of exam of this problem. We nevertheless take the additional problem of writing a curtailed argument of proof of the demonstration -- that the solution volition e'er have the 2 lines of slope thou and -m crossing on y = 1 and the vertex of the parabola on y = 1/two.

Each senario presents a somewhat different arroyo. Which would be most helpful in finding two quadratic functions f(10) and g(ten) such that their product part h(x) = f(x).k(10) has each tangent? The following graphs show such functions. How tin can they be generated?

What are some generalizations of the problem (and the solutions)?

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Source: http://jwilson.coe.uga.edu/Texts.Folder/tangent/f(x).g(x)%3Dh(x).html